[cygaril]

Assuming the definition of ~ is meant to be the same on Q and W (I think there's a typo), then it's no longer transitive and therefore no longer an equivalence relation: (0, 0) ~ (1, 2) and (0, 0) ~ (2, 3) but not (1, 2) ~ (2, 3).

[gizmo686]

I'm passed the window where I can edit my post, so I will correct it here.

You are correct that my original post is mistaken. I wish I could attribute that to a typo, but it was really me just working from memory without checking my work. A correct equivalence relation for a wheel over integers is as follows:

    (a,b) ~ (x,y) iff there exists s1,s2 (both non-zero integers) such that:
        (s1 * a, s1 * b) = (s2 * x, s2 * y)

When b,y are non zero, this is the same as the equivalence relationship on fractions.